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Prob3.5.6

# Prob3.5.6 - Stat 421 Solutions for Homework Set 5 Page 129...

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MATH 510-501 Probability and Statistics I; Fall 2003 Homework 6; Problem 3.5.6 Tony Smaldone Problem 3.5.6 Suppose that in a certain drug the concentration of a particular chemical is a random variable with a continuous distribution for which the p.d.f. is as follows: g ( x ) = 3 8 x 2 for 0 x 2 , 0 otherwise . Suppose that the concentrations X and Y of the chemical in two separate batches of the drug are independent random variables for each of which the p.d.f. is g . Determine a. the joint p.d.f. of X and Y b. P ( X = Y ) c. P ( X > Y ) d. P ( X + Y 1) Solution: a. Given that X and Y are independent, the following relationship is true f ( x, y ) = f X ( x ) f Y ( y ) . Since X and Y both have the p.d.f. g , then f ( x, y ) = g ( x ) g ( y ). Hence f ( x, y ) = 3 8 x 2 3 8 y 2 = 9 64 x 2 y 2 , 0 x 2 and 0 y 2 , 0 , otherwise . b. To evaluate P ( X = Y ) we need to define the region of integration. Here we have y = x , so the limits for y are x y x and 0 x 2 for x. Hence, P ( X = Y ) = 2 0 x x 9 64 x 2 y 2 dydx = 0 . As discussed in problem 3.4.4, the probability that X and Y will lie on a straight line is zero. c. To evaluate P ( X > Y ) we need to define the region of integration. Here we have y < x , so the limits for y are 0 y x and are 0 x 2 for x. Hence,

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